Exercise 40 Page 553

We want to find the area and the perimeter of the given quadrilateral. In order to do so, we first need to examine the given figure to find out the length of the unknown side.

Notice that we can split the figure into a rectangle and a right triangle. Recall the following theorem.

Parallelogram Opposite Sides Theorem
If a quadrilateral is a parallelogram, then its consecutive angles are supplementary.

Since rectangles are parallelograms, we can use that to find the lengths of the following segments.

Hypotenuse

To find the missing side of the triangle, we will use the Pythagorean Theorem.

a^{2} + b^{2} = c^{2}

In the formula,

a

and

b

are the legs and

c

is the hypotenuse of a right triangle. As we have found previously,

a = 6

and

b = 8 .

Let's substitute these values into the formula.

a^{2} + b^{2} = c^{2}

SubstituteII

$a = 6$ , $b = 8$

6^{2} + 8^{2} = c^{2}

▼

Solve for

c

CalcPow

Calculate power

36 + 64 = c^{2}

AddTerms

Add terms

100 = c^{2}

SqrtEqn

$LHS = RHS$

10 = c

RearrangeEqn

Rearrange equation

c = 10

Since a negative side length does not make sense, we only need to consider positive solutions.

Perimeter

Now that we have found out the hypotenuse of the right triangle, we can place its length in the given quadrilateral.

The perimeter of a figure is the sum of the side lengths.

P = 8 + 4 + 10 + 10 = 32

Therefore, the quadrilateral's perimeter is

32

units.

Area

Here we will calculate the areas of the rectangle and the right triangle separately. The total area is the sum of both parts.

Let's begin by calculating the area of the triangle.

A = \frac{1}{2} b h

Now we substitute the value of the base,

b = 6,

and the value of the height,

h = 8,

into the formula to calculate

A .

A = \frac{1}{2} b h

SubstituteII

$b = 6$ , $h = 8$

A = \frac{1}{2} (6) (8)

Multiply

A = \frac{1}{2} (48)

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

A = \frac{1 \cdot 4 8}{2}

Multiply

A = \frac{4 8}{2}

CalcQuot

Calculate quotient

A = 24

The area of the triangle is

24

square units. Now, let's calculate the area of the rectangle.

A = ℓ w

Now we substitute the length,

ℓ = 4,

and the width,

w = 8,

of the rectangle into the formula to calculate

A .

A = ℓ w

SubstituteII

$ℓ = 4$ , $w = 8$

A = 4 (8)

Multiply

A = 32

The area of the rectangle is

32

square units. Recall that the total area is the sum of the areas of the right triangle and the rectangle.

A_{T} = 24 + 32 = 56

Therefore, the total area of the figure is

56

square units.